The equation of the chord, of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$, whose mid-point is $(3,1)$ is :
- $48 x+25 y=169$
- $5 x+16 y=31$
- $25 x+101 y=176$
- $4 x+122 y=134$
Solution
$\begin{aligned}
& T=S_1 \\ & \Rightarrow \frac{3 x}{25}+\frac{\mathrm{y}}{16}-1=\frac{9}{25}+\frac{1}{16}-1 \\ & 48 \mathrm{x}+25 \mathrm{y}=144+25 \\ & 48 \mathrm{x}+25 \mathrm{y}=169
\end{aligned}$
Asked in: JEE Main 2025 (24 Jan Shift 2)